Question

The equation ${{x}^{2}}-6x+8+\lambda \left( {{x}^{2}}-4x+3 \right)=0\forall \lambda \in R-\left\{ -1 \right\}$ has:
(a). Real and unequal roots for all $\lambda \in R-\left\{ -1 \right\}$
(b). Real and equal roots for $\lambda <0$
(c). Real and equal roots for $\lambda >0$
(d). Real and equal roots for $\lambda =0$

Answer 1

Hint: Rewriting the given equation as ${{x}^{2}}\left( 1+\lambda \right)-\left( 6+4\lambda \right)x+8+3\lambda =0$. Now, write the discriminant value (D) for this quadratic equation. Discriminant value of this quadratic equation is ${{\left( -\left( 6+4\lambda \right) \right)}^{2}}-4\left( 1+\lambda \right)\left( 8+3\lambda \right)$. Now, see whether the value of D could be 0 or not. We know the property of the quadratic equation that when D is equal to 0 then roots are real and equal and when the value of D is not equal to 0 then roots are definitely not equal but could be real or imaginary.

Complete step-by-step answer:
The quadratic equation given in the question is:
 ${{x}^{2}}-6x+8+\lambda \left( {{x}^{2}}-4x+3 \right)=0\forall \lambda \in R-\left\{ -1 \right\}$
Rewriting the above equation we get,
${{x}^{2}}\left( 1+\lambda \right)-\left( 6+4\lambda \right)x+8+3\lambda =0$…………. Eq. (1)
To find the nature of the roots of the quadratic equation we will find the value of discriminant (or D).
The discriminant (or D) of the quadratic equation $a{{x}^{2}}+bx+c=0$ is:
$D={{b}^{2}}-4ac$
So, the value of D for eq. (1) is:
$\begin{align}
  & D={{\left( -\left( 6+4\lambda \right) \right)}^{2}}-4\left( 1+\lambda \right)\left( 8+3\lambda \right) \\
 & \Rightarrow D=36+16{{\lambda }^{2}}+48\lambda -4\left( 8+3\lambda +8\lambda +3{{\lambda }^{2}} \right) \\
\end{align}$
$\begin{align}
  & \Rightarrow D=36+16{{\lambda }^{2}}+48\lambda -32-44\lambda -12{{\lambda }^{2}} \\
 & \Rightarrow D=4{{\lambda }^{2}}+4\lambda +4 \\
\end{align}$
In the above equation taking 4 as common and writing the remaining expression in the bracket we get,
$D=4\left( {{\lambda }^{2}}+\lambda +1 \right)$………. Eq. (2)
It is given that $\lambda $ can take any value except -1.
It is a property in a quadratic equation that when the value of D is equal to 0 then the roots are real and equal and when the value of D is not equal to 0 then roots are definitely not equal but can be real or imaginary.
Rearranging the eq. (2) we get,
$D=4\left( {{\left( \lambda +\dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2} \right)$……. Eq. (3)
As you can see from the expression above, no matter whatsoever value of $\lambda $ you will put, the expression will never attain the value 0 so real and equal roots could not be possible.
From the above result, options (b), (c) and (d) will be cancelled and we are remaining with the option (a).
Hence, the correct option is (a).

Note: You might be wondering how we have come from eq. (2) to eq. (3). We are showing the answer below.
$D=4\left( {{\lambda }^{2}}+\lambda +1 \right)$
Make the perfect square of the quadratic expression written in the brackets by adding or subtracting the quadratic expression by $\dfrac{1}{2}$.
$D=4\left( {{\lambda }^{2}}+\lambda +\dfrac{1}{2}+1-\dfrac{1}{2} \right)$
From the above expression you can see that $\left( {{\lambda }^{2}}+\lambda +\dfrac{1}{2} \right)$ is the expansion of ${{\left( \lambda +\dfrac{1}{2} \right)}^{2}}$.
$D=4\left( {{\left( \lambda +\dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2} \right)$
The above equation is what we were getting as eq. (3) in the above solution.
Hence, we have shown how we have come from eq. (2) to eq. (3).